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3 years ago

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A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The

concentration of CaCl2 in this solution is ________ molal. A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is ________ molal. 5.70 0.214 0.569 63.2 1.76

1 answer:

polet [3.4K]3 years ago

3 0

Answer:

The concentration of CaCl2 in this solution is 0.563 M

Explanation:

<u>Step 1:</u> Data given

mass of CaCl2 = 23.7 grams

Molar mass of CaCl2 = 110.98 g/mol

mass of water = 375 grams

Molar mass of water = 18.02 grams

Volume of the solution = V

<u>Step 2:</u> Calculate the mass of solution

Mass of the solution = Mass of solute + Mass of solvent

Mass of the solution = 23.7 grams + 375 grams

Mass of the solution = 398.7 grams

<u>Step 3:</u> Calculate the volume

Volume = mass / density

Volume = 398.7 grams / 1.05 mg/mL

Volume = 379.71 mL = 0.3791 L

<u>Step 4:</u> Calculate moles of CaCl2

Number of moles CaCl2 = mass CaCl2 / Molar mass CaCl2

Number of moles CaCl2 = 23.7 grams / 110.98 g/mol

Number of moles CaCl2 = 0.21355 moles

<u>Step 5:</u> Calculate the concentration of CaCl2

Concentration = Number of moles / Volume

Concentration = 0.21355 moles / 0.3791 L

Concentration = 0.563 M

The concentration of CaCl2 in this solution is 0.563 M

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